考研851 自动控制原理
题海 · pdf-page · p.26
\[ e^{\boldsymbol{A}} = \boldsymbol{I} + \frac{\boldsymbol{A}}{1!} + \frac{\boldsymbol{A}^2}{2!} + \cdots + \frac{\boldsymbol{A}^k}{k!} + \cdots = \sum_{k=0}^{\infty} \frac{\boldsymbol{A}^k}{k!} \]

一个关于时间的矩阵指数函数定义为

\[ e^{\boldsymbol{A}t} = \sum_{k=0}^{\infty} \frac{\boldsymbol{A}^k t^k}{k!} \]

若矩阵指数函数对时间微分,则有

\[ \frac{\mathrm{d}}{\mathrm{d}t}(e^{\boldsymbol{A}t}) = \boldsymbol{A}e^{\boldsymbol{A}t} = e^{\boldsymbol{A}t}\boldsymbol{A} \]

3) 矩阵乘积的微分

如果矩阵 \(\boldsymbol{A}(t)\)\(\boldsymbol{B}(t)\)\(t\) 是可微的,则有

\[ \frac{\mathrm{d}}{\mathrm{d}t}[\boldsymbol{A}(t)\boldsymbol{B}(t)] = \frac{\mathrm{d}\boldsymbol{A}(t)}{\mathrm{d}t}\boldsymbol{B}(t) + \boldsymbol{A}(t)\frac{\mathrm{d}\boldsymbol{B}(t)}{\mathrm{d}t} \]

4) 逆矩阵的微分

如果矩阵 \(\boldsymbol{A}(t)\) 及其逆矩阵 \(\boldsymbol{A}^{-1}(t)\)\(t\) 是可微的,那么 \(\boldsymbol{A}^{-1}(t)\) 的微分可以导出如下等式:

\[ \frac{\mathrm{d}}{\mathrm{d}t}[\boldsymbol{A}(t)\boldsymbol{A}^{-1}(t)] = \frac{\mathrm{d}\boldsymbol{A}(t)}{\mathrm{d}t}\boldsymbol{A}^{-1}(t) + \boldsymbol{A}(t)\frac{\mathrm{d}\boldsymbol{A}^{-1}(t)}{\mathrm{d}t} \]

因为

\[ \frac{\mathrm{d}}{\mathrm{d}t}[\boldsymbol{A}(t)\boldsymbol{A}^{-1}(t)] = \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol{I} = \boldsymbol{0} \]

\[ \boldsymbol{A}(t)\frac{\mathrm{d}\boldsymbol{A}^{-1}(t)}{\mathrm{d}t} = -\frac{\mathrm{d}\boldsymbol{A}(t)}{\mathrm{d}t}\boldsymbol{A}^{-1}(t) \]

于是有

\[ \frac{\mathrm{d}\boldsymbol{A}^{-1}(t)}{\mathrm{d}t} = -\boldsymbol{A}^{-1}(t)\frac{\mathrm{d}\boldsymbol{A}(t)}{\mathrm{d}t}\boldsymbol{A}^{-1}(t) \]

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